📚 9630-PH02 2016 Mark Scheme: Deriving Resistors in Parallel | 9630-PH02 2016 评分标准：并联电阻公式推导
In Edexcel International AS Physics Unit 2 (PH02), candidates are frequently asked to derive the formula for the total resistance of two resistors connected in parallel. The 2016 mark scheme provides a clear framework of the logical steps expected. This article walks through that derivation in detail, highlighting the key points that earn marks and the common errors students should avoid.
在爱德思国际AS物理单元2（PH02）中，考生经常被要求推导两个并联电阻的总电阻公式。2016年的评分标准提供了清晰的逻辑步骤框架。本文将详细讲解该推导过程，突出得分关键点，并指出学生需要避免的常见错误。
1. The Derivation Question in PH02 | PH02 中的推导题
The derivation typically states: “Use the principles of current conservation and Ohm’s law to show that, for two resistors R₁ and R₂ connected in parallel, the total resistance R is given by 1/R = 1/R₁ + 1/R₂.” The mark scheme rewards a systematic, step-by-step approach with clear algebraic justification.
推导题通常这样陈述：“利用电流守恒原理和欧姆定律证明，对于并联的两个电阻R₁和R₂，总电阻R由公式1/R = 1/R₁ + 1/R₂给出。”评分标准奖励系统化、分步推导，并要求清晰的代数论证。
2. Essential Electrical Principles | 关键电学原理
Before beginning the derivation, recall the two fundamental rules that apply to any parallel circuit: the total current entering a junction equals the sum of the currents in each branch, and the potential difference across all parallel branches is the same.
在开始推导之前，请回顾适用于任何并联电路的两条基本原则：流入节点的总电流等于各支路电流之和，且所有并联支路两端的电位差相同。
These principles are rooted in charge conservation and energy conservation, respectively. In the mark scheme, simply stating these principles correctly can earn the first available marks.
这些原理分别植根于电荷守恒和能量守恒。在评分标准中，仅正确陈述这些原理就能获得最初的分数。
3. Step 1: Current Conservation | 步骤1：电流守恒
Let the current from the source be I. At the junction where the circuit splits into the two parallel branches, the current divides into I₁ through resistor R₁ and I₂ through resistor R₂. By charge conservation, the total current I is the sum of these branch currents.
设电源提供的电流为I。在电路分叉为两条并联支路的节点处，电流分为流过电阻R₁的I₁和流过电阻R₂的I₂。根据电荷守恒，总电流I等于这些支路电流之和。
I = I₁ + I₂
It is essential to assign clear labels to all currents and to write the sum explicitly. Mark schemes often award a mark for this starting equation.
必须为所有电流设置清晰的标记，并明确写出求和等式。评分标准通常为此起始方程分配一个分数。
4. Step 2: Equal Voltage Across Parallel Branches | 步骤2：并联支路电压相等
Since the resistors are connected in parallel, the same potential difference V appears across both R₁ and R₂. This is because each resistor is connected directly to the same two points in the circuit. We can write:
由于电阻是并联的，相同的电位差V同时出现在R₁和R₂两端。这是因为每个电阻都直接连接在电路中的相同两个点之间。我们可以写出：
V = V₁ = V₂
Many students lose a mark here by either not stating this equality explicitly or by assuming different voltages for each branch. The mark scheme expects a clear statement that the p.d. is common to both resistors.
许多学生在此失分，原因是没有明确陈述这一相等关系，或者假设了各支路电压不同。评分标准期望明确指出两个电阻的端电压相同。
5. Step 3: Applying Ohm’s Law to Each Resistor | 步骤3：对各电阻应用欧姆定律
Ohm’s law relates the current through a resistor to the potential difference across it. For each resistor, we can express the current as V divided by its resistance. Apply this to both resistors individually.
欧姆定律将流过电阻的电流与其两端的电位差联系起来。对于每个电阻，我们可以将电流表示为电压除以电阻。将这一关系分别应用于两个电阻。
I₁ = V / R₁ and I₂ = V / R₂
At this point, the derivation begins to take shape algebraically. Using the same symbol V for both equations is correct because we have already established that the voltage is the same.
此时，推导在代数上开始成形。两个方程使用相同的符号V是正确的，因为我们已经确定了电压相同。
6. Step 4: Substituting into the Total Current Equation | 步骤4：代入总电流方程
Take the current conservation equation I = I₁ + I₂ and replace I₁ and I₂ with the expressions from Ohm’s law. This yields an equation that involves the total current I, the common voltage V, and the individual resistances.
取电流守恒方程I = I₁ + I₂，并用欧姆定律表达式替换I₁和I₂。这样就得到了一个包含总电流I、公共电压V和各个电阻的方程。
I = V / R₁ + V / R₂
The mark scheme rewards this substitution step because it demonstrates the candidate can link the physical principles together. Be meticulous with parentheses if needed, though here the expression is simple.
评分标准奖励这一代入步骤，因为它表明考生能够将物理原理联系起来。尽管这里的表达式很简单，但如果需要，仍应仔细使用括号。
7. Step 5: Simplifying to Obtain the Reciprocal Formula | 步骤5：简化得到倒数公式
The total resistance R of the parallel combination is defined as the ratio of the total voltage V to the total current I, i.e. R = V / I. Rearrange the equation to express V/I in terms of R₁ and R₂.
并联组合的总电阻R定义为总电压V与总电流I之比，即R = V / I。重新整理方程，用R₁和R₂表示V/I。
From I = V(1/R₁ + 1/R₂), divide both sides by V (assuming V ≠ 0) to obtain:
由I = V(1/R₁ + 1/R₂)，两边同时除以V（假设V ≠ 0），得到：
I / V = 1/R₁ + 1/R₂
Since R = V / I, the reciprocal of the total resistance is I / V. Therefore, we arrive at the final derived formula.
因为R = V / I，总电阻的倒数就是I / V。因此，我们得到最终的推导公式。
1 / R = 1 / R₁ + 1 / R₂
Mark schemes specifically look for the correct manipulation and the final expression. Some even award a separate mark for stating that this result holds for any number of resistors in parallel, but the 2016 question likely focused on two resistors.
评分标准特别看中正确的代数操作和最终表达式。有些甚至单独给分，用于陈述这一结果适用于任意数量的并联电阻，但2016年的题目很可能只关注两个电阻。
8. Checking the Mark Scheme Points | 核对评分要点
Review the typical distribution of marks in a 3- or 4-mark derivation: one mark for stating I = I₁ + I₂, one for recognizing V is the same across both branches, one for substituting I₁ = V/R₁ and I₂ = V/R₂, and one for correctly rearranging to 1/R = 1/R₁ + 1/R₂. Some schemes also award a mark for explicitly stating the definition of total resistance.
回顾一个3分或4分推导题的典型分数分配：一个分数用于写出I = I₁ + I₂，一个分数用于认识到V在两支路相等，一个分数用于代入I₁ = V/R₁和I₂ = V/R₂，一个分数用于正确整理得到1/R = 1/R₁ + 1/R₂。有些评分标准还给分用于明确陈述总电阻的定义。
To maximize marks, present each logical step on a new line with a short justification. Phrases like “by charge conservation”, “since p.d. is the same”, and “by Ohm’s law” help the examiner award marks.
为获得最高分，应将每个逻辑步骤另起一行，并附上简短的依据。诸如“由电荷守恒”、“因为电位差相同”和“根据欧姆定律”等短语有助于考官给分。
9. Common Pitfalls and How to Avoid Them | 常见陷阱与避免方法
One frequent mistake is writing I₁ = R₁ / V or confusing the reciprocal relationship. Always double-check that Ohm’s law is expressed correctly as I = V/R, not R/V. Another common error is forgetting to state that the voltage V is the same for both branches before substitution.
一个常见错误是将I₁写成R₁ / V，或者混淆了倒数关系。务必复查欧姆定律是否正确地表达为I = V/R，而非R/V。另一个常见错误是在代入之前忘记陈述两支路电压V相等。
Some students try to use the formula for series resistors by mistake, so keep the circuit diagram clear in your mind. Additionally, when manipulating the final equation, ensure you do not incorrectly write R = R₁ + R₂ for parallel resistors. Always end with the reciprocal form.
有些学生错误地尝试使用串联电阻公式，因此请牢记电路图。此外，在整理最终方程时，确保不会错误地将并联电阻写成R = R₁ + R₂。始终以倒数形式结束。
Also, note that the mark scheme may penalize the omission of the condition that V ≠ 0, though this is usually implied.
还需注意，评分标准可能会因遗漏V ≠ 0的条件而扣分，不过这通常是隐含的。
10. Practice with Example Values | 实例练习
Test your derived formula: suppose R₁ = 6 Ω and R₂ = 3 Ω. Using 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = ½, so R = 2 Ω. This total is less than either individual resistance, which is a characteristic of parallel circuits.
检验你推导出的公式：假设R₁ = 6 Ω，R₂ = 3 Ω。由1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = ½，得R = 2 Ω。这一总值比任何一个单独电阻都小，这是并联电路的一个特征。
Being able to check numerical consistency gives you confidence in exam conditions. If your calculated R is larger than the smallest resistor, you have likely made an error.
能够通过数值检验一致性可以在考试环境中给你信心。如果你算出的R比最小的电阻还要大，那么很可能出错了。
11. Applying the Derivation to More Complex Circuits | 将推导应用于更复杂电路
Once you master the two‑resistor derivation, you can easily extend it to three or more parallel resistors. The same logic yields 1/R = 1/R₁ + 1/R₂ + 1/R₃ + …, which the mark scheme may reward if you justify using the identical principles.
一旦掌握了两个电阻的推导，就能轻松将其扩展到三个或更多并联电阻。相同的逻辑会得到1/R = 1/R₁ + 1/R₂ + 1/R₃ + …，如果你用相同的原理加以论证，评分标准也可能给分。
This derivation also underpins the famous “product over sum” rule for two resistors: R = (R₁ R₂) / (R₁ + R₂). This is obtained by taking the reciprocal of both sides after finding a common denominator. It is a useful shortcut, but in a “show that” question, always present the full step‑by‑step derivation.
该推导也支撑着著名的两个电阻“积比和”规则：R = (R₁ R₂) / (R₁ + R₂)。这是通过通分后取倒数得到的。虽然这是一个有用的快捷方式，但在“证明”类题目中，务必给出完整的分步推导。
12. Conclusion | 结论
The derivation of the parallel resistance formula from first principles is a straightforward yet highly examination‑relevant skill. By following the structured approach shown in the 2016 mark scheme — stating conservation laws, using Ohm’s law, and performing careful algebra — you can secure all available marks. Practice writing out the derivation under timed conditions until it becomes second nature.
从基本原理推导并联电阻公式是一项直接但考试相关性极高的技能。通过遵循2016年评分标准所示的结构化方法——陈述守恒定律、使用欧姆定律并进行细致的代数运算——你可以确保获得所有可能的分数。请在限时条件下反复练习写出该推导，直到它成为你的本能。
Remember, clarity of reasoning is as important as the final formula itself. Each step should be accompanied by a brief physical justification, exactly as the examiners expect.
记住，推理的清晰度与最终公式本身同样重要。每一步都应附上简短的物理依据，这恰恰是考官所期望的。
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